1. Tardigrade
  2. Question
  3. Mathematics
  4. i2+i4+i6+...(2n+1) terms =

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. $i^{2}+i^{4}+i^{6}+...(2n+1)$ terms =

Complex Numbers and Quadratic Equations

Solution:

The first $2n$ terms will cancel in pairs and the last term is
$\left(i^{2}\right)^{2n+1} =i^{4n} \cdot i^{2}$
$=-1$